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Chapter 3: Q 3.5-7E (page 121)

An industrial electromagnet can be modeled as an RLcircuit, while it is being energized with a voltage source. If the inductance is 10 H and the wire windings contain of resistance, how long does it take a constant applied voltage to energize the electromagnet to within 90% of its final value (that is, the current equals 90% of its asymptotic value)?

Short Answer

Expert verified

The value or time is 7.68 sec.

Step by step solution

01

Determine the value of I(t)

Here given R=3Ω,L=10H

We have the equation

I(t)=e-RLt(∫eRLVLdt+K)I(t)=e-310t(∫e310V10dt+K)I(t)=e-310t(∫e310V10dt+K)I(t)=e-310t(e310tV3+K)I(t)=V3+Ke-310t

Apply the given condition then we get the value, k=-v3

I(t)=V3+(1-e-310t).

02

Apply the asymptotic value on the equation

limt→∞lt=limt→∞V3+1-e-310t=V3

03

Evaluate the value of time

Since the 90% 0f the asymptotic value of current is 0.9V3.

I(t)=0.9V3I(t)=V3(1-e-310t)=0.9V3(1-e-310t)=0.9

By solving for t = 7.68 sec.

Hence, the value or time is 7.68 sec.

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